Concise Mathematics Selina Solutions Class 10 ICSE Chapter 4 Linear Inequations (In one variable)

This page provides the most comprehensive ICSE Class 10 Maths Linear Inequations Solutions, covering Chapter 4 from the Selina Concise Mathematics textbook. Unlike equations which have a single solution, linear inequations introduce the concept of a solution set – a range of values that satisfy the condition. This chapter is crucial for building a strong foundation in algebra, as you will learn to solve for variables using inequality symbols like <, >, ≤, and ≥. We will explore how the solution changes based on the replacement set, whether it’s Natural numbers (N), Whole numbers (W), Integers (Z), or Real numbers (R), and master the skill of representing these solution sets on a number line.

If you are stuck on a particular question or want to verify your method for solving inequations, you have come to the right place. This page contains detailed, step-by-step solutions for all 56 questions found in Exercise 4(A), Exercise 4(B), the Test Yourself section, and the Case Study Based Question. Each solution is crafted to follow the exact methodology and presentation style expected by the ICSE board, ensuring you learn the correct way to approach these problems for your exams. Here, you will find clear, accurate, and easy-to-follow solutions to master linear inequations.

Exercise 4(A)

Question 1(a)

If x ∈ W, then solution set of inequation -x > -7, is :

• (a) {8, 9, 10, …..}
• (b) {0, 1, 2, 3, 4, 5, 6}
• (c) {0, 1, 2, 3, ….}
• (d) {-8, -9, -10, ….}

We start with the inequality:

By multiplying both sides by -1, we must remember to reverse the inequality sign. This gives us:

Since we are dealing with whole numbers (x \in W), the possible values for x are all whole numbers less than 7. Therefore, the solution set is:

Hence, Option 2 is the correct option.

Question 1(b)

The value of x, for 4(2x – 5) < 2x + 28, x ∈ R, is :

• (a) x > 8
• (b) x < 8
• (c) x > -8
• (d) x < -8

We start with the inequality:

⇒ 4(2x – 5) < 2x + 28

Distribute the 4 on the left side:

⇒ 8x – 20 < 2x + 28

To isolate the terms with x, subtract 2x from both sides:

⇒ 8x – 2x < 28 + 20

Simplify the inequality:

⇒ 6x < 48

Divide both sides by 6 to solve for x:

⇒ x < \dfrac{48}{6}

Simplifying gives:

⇒ x < 8.

Hence, Option 2 is the correct option.

Question 1(c)

The solution set for the inequation -2x + 7 ≤ 3, x ∈ R is :

• (a) {x : x ∈ R, x < 2}
• (b) {x : x ∈ R, x > 2}
• (c) {x : x ∈ R, x ≤ 2}
• (d) {x : x ∈ R, x ≥ 2}

We start with the inequality:

⇒ -2x + 7 ≤ 3

First, subtract 7 from both sides:

⇒ -2x ≤ 3 – 7

Simplifying the right side, we get:

⇒ -2x ≤ -4

Now, divide each side by -2. Remember, dividing by a negative number reverses the inequality sign:

⇒ x ≥ \dfrac{-4}{-2}

⇒ x ≥ 2

Since x is a member of the real numbers (x ∈ R), the solution set is:

{x : x ∈ R, x ≥ 2}

Hence, Option 4 is the correct option.

Question 1(d)

For 7 – 3x < x – 5, the solution set is :

• (a) x > 3
• (b) x < 3
• (c) x ≥ 3
• (d) x ≤ 3

Consider the inequality:

⇒ 7 – 3x < x – 5

First, let’s bring all terms involving x to one side and constant terms to the other. Add 3x to both sides:

⇒ 7 < x + 3x – 5

This simplifies to:

⇒ 7 < 4x – 5

Now, add 5 to both sides to isolate terms involving x:

⇒ 7 + 5 < 4x

⇒ 12 < 4x

Next, divide both sides by 4 to solve for x:

⇒ x > \dfrac{12}{4}

⇒ x > 3

Hence, Option 1 is the correct option.

Question 1(e)

x(8 – x) > 0 and x ∈ N gives :

• (a) 0 ≤ x < 8
• (b) 1 < x ≤ 8
• (c) 0 < x < 8
• (d) 0 ≤ x ≤ 8

Consider the inequality x(8 – x) > 0.

We have two possible scenarios:

1. When both factors are positive:
2. x > 0
3. 8 – x > 0 ⇒ x < 8

Therefore, combining these, we get 0 < x < 8.

1. When both factors are negative:
2. x < 0
3. 8 – x < 0 ⇒ x > 8

Notice that these conditions cannot occur at the same time, as x cannot be both less than 0 and greater than 8 simultaneously.

Thus, the only valid solution set is from the first scenario: {0 < x < 8}.

Hence, Option 3 is the correct option.

Question 2

State, true or false :

(i) x < -y ⇒ -x > y

(ii) -5x ≥ 15 ⇒ x ≥ -3

(iii) 2x ≤ -7 ⇒ \dfrac{2x}{-4} \ge \dfrac{-7}{-4}

(iv) 7 > 5 ⇒ \dfrac{1}{7} \lt \dfrac{1}{5}

(i) We start with the inequality:

x < -y

Applying the rule for inequalities when multiplying or dividing by a negative number, we reverse the inequality sign:

∴ -x > y

Thus, the statement is True.

(ii) Starting with:

-5x ≥ 15

By dividing each side by -5, we must flip the inequality sign:

⇒ x ≤ -3

This makes the statement False.

(iii) The given inequality is:

2x ≤ -7

Dividing both sides by -4, which is negative, requires us to reverse the inequality sign:

⇒ \dfrac{2x}{-4} \ge \dfrac{-7}{-4}

Thus, this statement is True.

(iv) We have:

7 > 5

On taking reciprocals, the inequality sign is flipped:

Hence, the statement is True.

Question 3

State, whether the following statements are true or false.

(i) If a < b, then a – c < b – c

(ii) If a > b, then a + c > b + c

(iii) If a < b, then ac > bc

(iv) If a > b, then \dfrac{a}{c} \lt \dfrac{b}{c}

(v) If a – c > b – d; then a + d > b + c

(vi) If a < b, and c > 0, then a – c > b – c

where a, b, c, and d are real numbers c ≠ 0.

(i) Start with the inequality a < b. By subtracting c from both sides, we maintain the inequality: a - c < b - c. Thus, this statement is True.

(ii) With a > b, adding c to both sides gives a + c > b + c. Therefore, this statement is True.

(iii) When a < b, consider two cases for c:
– If c is positive, multiplying by c yields ac < bc.
– If c is negative, multiplying by c reverses the inequality: ac > bc.
Thus, the statement is False.

(iv) Given a > b, consider:
– For positive c, dividing by c gives \dfrac{a}{c} > \dfrac{b}{c}.
– For negative c, dividing by c reverses the inequality: \dfrac{a}{c} < \dfrac{b}{c}.
So, the statement is False.

(v) From a - c > b - d, add (c + d) to both sides:
⇒ a - c + (c + d) > b - d + (c + d)
⇒ a + d > b + c.
Hence, this statement is True.

(vi) Given a < b and c > 0, subtracting c from both sides results in a - c < b - c. Thus, this statement is False.

Question 4

Solve the inequation :

3 – 2x ≥ x – 12 given that x ∈ N.

We start with the inequality:

⇒ 3 – 2x ≥ x – 12

To simplify, add 2x to both sides:

⇒ 3 ≥ x + 2x – 12

This simplifies to:

⇒ 3 ≥ 3x – 12

Next, add 12 to both sides to isolate terms involving x:

⇒ 3 + 12 ≥ 3x

⇒ 15 ≥ 3x

Now, divide both sides by 3 to solve for x:

⇒ 5 ≥ x

This can be rewritten as:

⇒ x ≤ 5

Since x belongs to the set of natural numbers (x ∈ N), the possible values for x are natural numbers less than or equal to 5.

∴ Solution set = {1, 2, 3, 4, 5}.

Question 5

If 25 – 4x ≤ 16, find :

(i) the smallest value of x when x is a real number,

(ii) the smallest value of x when x is an integer.

Let’s start with the inequality:

Subtract 25 from both sides:

This simplifies to:

By multiplying both sides by -1, remember to reverse the inequality sign:

Now, divide both sides by 4:

which is equivalent to:

(i) For real numbers:

Since x \geq 2.25, the smallest value of x is 2.25.

(ii) For integers:

Given x \geq 2.25, the smallest integer satisfying the inequality is 3.

Question 6

If the replacement set is the set of real numbers, solve :

(i) -4x ≥ -16

(ii) 8 – 3x ≤ 20

(i) Consider the inequality -4x \ge -16.

To isolate x, divide both sides by -4. Remember, dividing by a negative number reverses the inequality sign:

This simplifies to:

∴ The solution set is \{ x : x \in \mathbb{R} \text{ and } x \le 4 \}.

(ii) Now, examine the inequality 8 - 3x \le 20.

Subtract 8 from both sides to simplify:

This results in:

Next, divide both sides by -3. Keep in mind, dividing by a negative flips the inequality:

∴ The solution set is \{ x : x \in \mathbb{R} \text{ and } x \ge -4 \}.

Question 7

Find the smallest value of x for which 5 – 2x < 5\dfrac{1}{2} - \dfrac{5}{3}x, where x is an integer.

We start with the inequality:

This can be rewritten as:

Subtract \dfrac{11}{2} from both sides:

Simplifying the left side gives:

This simplifies to:

Multiply both sides by 3 to solve for x:

Which is the same as:

Since x must be an integer, the smallest integer greater than -1.5 is -1.

Hence, smallest value of x = -1.

Question 8

Find the largest value of x for which

2(x – 1) ≤ 9 – x and x ∈ W.

We start with the inequality:

⇒ 2(x – 1) ≤ 9 – x

Expanding the left side gives:

⇒ 2x – 2 ≤ 9 – x

Adding x to both sides results in:

⇒ 2x + x ≤ 9 + 2

Simplifying further, we have:

⇒ 3x ≤ 11

Dividing both sides by 3, we find:

⇒ x ≤ \dfrac{11}{3}

⇒ x ≤ 3.67

Since x is a whole number (x ∈ W), the largest integer value satisfying the inequality is 3.

Hence, largest value of x for which 2(x – 1) ≤ 9 – x is 3.

Exercise 4(B)

Question 1(a)

For the following real number line, the solution set is :

• (a) {x : x ∈ W and -2 ≤ x < 4}
• (b) {x : x ∈ N and -2 ≤ x ≤ -4}
• (c) {x : x ∈ R and -2 < x ≤ 4}
• (d) {x : x ∈ R and -2 ≤ x < 4}

Observing the real number line, we can determine that the value of x belongs to the set of real numbers R, where x is greater than -2 and less than or equal to 4.

∴ The solution set is expressed as \{ x : x \in R \text{ and } -2 < x \leq 4 \}.

Option 3 is the correct option.

Question 1(b)

The solution set for the following number line is :

• (a) {x : x ∈ Z and -3 < x < 4}
• (b) {x : x ∈ Z and -3 ≤ x}
• (c) {x : x ∈ Z and -2 ≤ x ≤ 4}
• (d) {x : x ∈ Z and -3 ≤ x ≤ 4}

Observing the number line, we see that the values of x are all integers, denoted by x \in \mathbb{Z}, and these values start from -3 and extend to the right. This indicates that x is greater than or equal to -3. Therefore, the solution set can be expressed as \{x : x \in \mathbb{Z} \text{ and } -3 \leq x\}.

Hence, option 2 is the correct option.

Question 1(c)

The following number line represents :

• (a) {x : x ∈ R and x = 10}
• (b) {(x < 10) ∪ (x > 10)}
• (c) {(10 > x) ∩ (x > 10)}
• (d) {x : x ∈ R and x < 10}

Observe the number line provided. It clearly indicates that the solution set is composed of all real numbers except 10. This can be expressed as the union of two sets: one where x is less than 10 and another where x is greater than 10. Thus, the solution set is represented as ((x < 10) \cup (x > 10)).

Hence, Option 2 is the correct option.

Question 1(d)

The solution set for the following number line is :

• (a) {x : x ∈ R, x < -2 and x > 3}
• (b) {x : x ∈ R and -2 < x < 3}
• (c) {x : x ∈ R, x < -2 or x < 3}
• (d) {x : x ∈ R, x ≤ -2 or x ≥ 3}

Observing the given number line, we identify the solution set as follows:

The values of x belong to the set of real numbers R, such that x is less than or equal to -2 or x is greater than or equal to 3.

∴ The solution set is expressed as \{ x : x \in R, x \leq -2 \text{ or } x \geq 3 \}.

Hence, option 4 is the correct option.

Question 1(e)

The number line for the solution of inequation x > 5 and x < 10 (x ∈ R) is :

Consider the conditions given: x > 5 and x < 10, where x belongs to the set of real numbers (x \in \mathbb{R}).

The solution set can be expressed as \{ x : x \in \mathbb{R} \text{ and } 5 < x < 10 \}.

Hence, Option 2 is the correct option.

Question 2

For each graph given alongside, write an inequation taking x as the variable :

(i) Observing the graph, we determine:

x ≤ -1 and x ∈ R.

(ii) By analyzing the graph, we conclude:

x ≥ 2 and x ∈ R.

(iii) Interpreting the graph, it shows:

-4 ≤ x < 3 and x ∈ R.

(iv) Examining the graph, we find:

-1 < x ≤ 5 and x ∈ R.

Question 3(i)

For the following inequations, graph the solution set on the real number line :

-4 ≤ 3x – 1 < 8

Consider the inequality:

-4 ≤ 3x – 1 < 8

Let’s handle the left-hand side first:

⇒ -4 ≤ 3x – 1

Adding 1 to both sides, we have:

⇒ 3x ≥ -4 + 1

⇒ 3x ≥ -3

Dividing by 3, we find:

⇒ x ≥ -1 \quad \text{…(i)}

Now, let’s solve the right-hand side:

⇒ 3x – 1 < 8

Adding 1 to both sides gives us:

⇒ 3x < 8 + 1

⇒ 3x < 9

Dividing by 3, we get:

⇒ x < 3 \quad \text{…(ii)}

Combining the results from (i) and (ii), we have:

-1 ≤ x < 3.

∴ Solution set = {-1 ≤ x < 3 : x ∈ \mathbb{R} }

The graphical representation on the number line is shown below:

Question 3(ii)

For the following inequations, graph the solution set on the real number line :

-1 < 3 – 2x ≤ 7

Consider the given inequality:

-1 < 3 – 2x ≤ 7

Let’s first tackle the left-hand side:

⇒ -1 < 3 – 2x

Subtract 3 from both sides:

⇒ -1 – 3 < -2x

⇒ -4 < -2x

Now, divide each side by -2, remembering to flip the inequality sign:

⇒ 2 > x

This can be rewritten as:

⇒ x < 2 \quad \text{…(i)}

Next, address the right-hand side:

⇒ 3 – 2x ≤ 7

Subtract 3 from both sides:

⇒ 3 – 7 ≤ 2x

⇒ -4 ≤ 2x

Divide both sides by 2:

⇒ -2 ≤ x

This gives us:

⇒ x ≥ -2 \quad \text{…(ii)}

Combining the results from (i) and (ii), we have:

-2 ≤ x < 2

∴ Solution set = {x : x ∈ \mathbb{R} \text{ and } -2 ≤ x < 2}

Solution on the number line is :

Question 3(iii)

For the following inequations, graph the solution set on the real number line :

x – 1 < 3 – x ≤ 5

Consider the inequality:

Step 1: Let’s address the left part of the inequality, x - 1 < 3 - x.

Add x to both sides:

This simplifies to:

Dividing both sides by 2, we find:

Step 2: Now, solve the right part, 3 - x \leq 5.

Rearranging gives:

Which simplifies to:

Combining Results: From these, we have two conditions:

1. x < 2
2. x \geq -2

These combine to form:

∴ Solution set = {-2 \leq x < 2 : x \in \mathbb{R}}

This solution set can be represented on the number line as follows:

!x - 1 < 3 - x \leq 5. Graph the solution set on the real number line. Linear Inequations, Concise Mathematics Solutions ICSE Class 10.(https://www.icseboard.org/images/q3ii-c4-ex-4-b-inequations-concise-maths-solutions-icse-class-10-1200×116.png)

Question 4

List the elements of the solution set of inequation -3 < x – 2 ≤ 9 – 2x; x ∈ N.

We start with the inequality:

⇒ -3 < x – 2 ≤ 9 – 2x

Let’s tackle the left side first:

⇒ -3 < x – 2

Adding 2 to both sides gives us:

⇒ x > -1 \quad \text{…….(i)}

Next, we’ll solve the right side:

⇒ x – 2 ≤ 9 – 2x

Rearranging terms, we have:

⇒ x + 2x ≤ 9 + 2

⇒ 3x ≤ 11

Dividing both sides by 3, we find:

⇒ x ≤ \dfrac{11}{3} \quad \text{………(ii)}

Combining the results from (i) and (ii), we get:

⇒ -1 < x ≤ \dfrac{11}{3}.

Given that x is a natural number (x ∈ N), we identify the natural numbers within this range:

∴ Solution set = {1, 2, 3}.

Question 5

Find the range of values of x which satisfies

-2\dfrac{2}{3} \le x + \dfrac{1}{3} \lt 3\dfrac{1}{3}, x ∈ R.

Graph these values on number line.

We start with the inequality:

First, convert the mixed numbers to improper fractions:

Let’s address the left-hand side of the inequality:

Subtract \dfrac{1}{3} from both sides:

Simplify the right side:

Which gives:

Next, solve the right-hand side of the inequality:

Subtract \dfrac{1}{3} from both sides:

Simplify the right side:

Which results in:

Combining results from (i) and (ii), we find:

Thus, the solution set is:

Solution set = {x : x \in R \text{ and } -3 \le x < 3}.

The solution on the number line is depicted as follows:

Question 6

Find the values of x, which satisfy the inequation :

-2 \le \dfrac{1}{2} - \dfrac{2x}{3} \le 1\dfrac{5}{6}, x ∈ N.

Graph the solution on the number line.

We start with the given inequality:

This can be rewritten as:

Let’s tackle the left side first:

Multiplying through by 6 to clear the fraction:

Rearranging gives:

Simplifying further, we find:

Now, let’s handle the right side:

Rearranging terms, we have:

This simplifies to:

Which further simplifies to:

Multiplying through by \dfrac{3}{2} gives:

Combining results from (i) and (ii), we obtain:

Since x \in N, the natural numbers that satisfy this inequality are:

∴ Solution set = {1, 2, 3}.

The solution is represented on the number line as shown in the image.

Question 7

Given x ∈ {real numbers}, find the range of values of x for which -5 ≤ 2x – 3 < x + 2 and represent it on a real number line.

We have the inequality:

Let’s tackle the left side of the inequality first:

Adding 3 to both sides results in:

Dividing every term by 2, we obtain:

Now, consider the right side of the inequality:

Subtracting x from both sides gives:

Combining the results from (i) and (ii), we find:

∴ The solution set is \{x : x \in \mathbb{R} \text{ and } -1 \leq x < 5\}.

The representation on the number line is shown below:

Question 8

If 5x – 3 ≤ 5 + 3x ≤ 4x + 2, express it as a ≤ x ≤ b and then state the values of a and b.

We start with the inequality:

⇒ 5x – 3 ≤ 5 + 3x ≤ 4x + 2

First, let’s handle the left part of the inequality:

⇒ 5x – 3 ≤ 5 + 3x

Rearranging terms, we get:

⇒ 5x – 3x ≤ 5 + 3

⇒ 2x ≤ 8

⇒ x ≤ 4 ……..(i)

Next, we solve the right part of the inequality:

⇒ 5 + 3x ≤ 4x + 2

Rearranging terms, we have:

⇒ 3x – 4x ≤ 2 – 5

⇒ -x ≤ -3

⇒ x ≥ 3 …….(ii)

Combining the results from (i) and (ii), we find:

3 ≤ x ≤ 4

Comparing this with the standard form a ≤ x ≤ b, we identify:

a = 3 and b = 4.

Hence, a = 3 and b = 4.

Question 9

Solve the following inequation and graph the solution set on the number line :

2x – 3 < x + 2 ≤ 3x + 5; x ∈ R.

We start with the compound inequality:

2x – 3 < x + 2 ≤ 3x + 5

Step 1: Solve the left part

For 2x – 3 < x + 2:

⇒ 2x – x < 2 + 3

⇒ x < 5 ( \text{………..(i)} )

Step 2: Solve the right part

For x + 2 ≤ 3x + 5:

⇒ x – 3x ≤ 5 – 2

⇒ -2x ≤ 3

⇒ 2x ≥ -3

Dividing both sides by 2:

⇒ x ≥ -1.5 ( \text{………(ii)} )

Combine the results

From (i) and (ii), we find:

-1.5 ≤ x < 5

∴ Solution set = {x : x ∈ R and -1.5 ≤ x < 5}.

The solution can be illustrated on the number line as shown in the image:

Question 10(i)

Solve and graph the solution set of :

2x – 9 < 7 and 3x + 9 ≤ 25; x ∈ R.

Consider the inequalities:

Let’s solve the first inequality:

Adding 9 to both sides gives:

Dividing each side by 2, we find:

Now, solving the second inequality:

Subtracting 9 from both sides results in:

Dividing each side by 3, we have:

From inequalities (i) and (ii), we conclude:

∴ Solution set = {x : x \leq 5\dfrac{1}{3} \text{ and } x \in \mathbb{R}}.

The solution is illustrated on the number line below:

Question 10(ii)

Solve and graph the solution set of :

2x – 9 ≤ 7 and 3x + 9 > 25; x ∈ I

Consider the inequations:

First, tackle the inequality 2x - 9 \leq 7:

Let’s label this as equation (i).

Next, address the inequality 3x + 9 > 25:

This simplifies to x > 5\dfrac{1}{3}. Let’s call this equation (ii).

Combining equations (i) and (ii), we find:

Since x \in I, which means x is an integer, the possible values for x are:

∴ Solution set = {6, 7, 8}.

Solution on the number line is :

Question 10(iii)

Solve and graph the solution set of :

x + 5 ≥ 4(x – 1) and 3 – 2x < -7; x ∈ R.

Consider the inequalities:

Let’s solve the first inequality:

Expanding the right side, we have:

Rearranging terms gives:

Dividing both sides by 3, we find:

Now, let’s solve the second inequality:

Adding 2x to both sides, we get:

Dividing by 2, we obtain:

From inequalities (i) and (ii), we have:

Notice that no number can simultaneously be less than or equal to 3 and greater than 5. ∴ the solution set is empty.

Hence, solution set is an empty set.

Question 11

Solve and graph the solution set of :

3x – 2 > 19 or 3 – 2x ≥ -7; x ∈ R

Consider the inequalities:

For the first inequality:

Add 2 to both sides:

This simplifies to:

Dividing both sides by 3 gives:

For the second inequality:

Add 2x to both sides:

Add 7 to both sides:

Simplifying, we get:

Dividing both sides by 2 results in:

Thus, the solution set is:

This solution can be represented on a number line, where values greater than 7 are on one side and values less than or equal to 5 are on the other.

Question 12

The diagram represents two inequations A and B on a real number lines :

(i) Write down A and B in set builder notation.

(ii) Represent A ∩ B and A ∩ B’ on two different number lines.

(i) For the inequality A, we can express it in set builder notation as follows: A = {x : -2 \leq x < 5 \text{ and } x \in \mathbb{R}}. This means x is a real number that satisfies -2 \leq x < 5.

Similarly, for the inequality B, it is written as: B = {x : -4 \leq x < 3 \text{ and } x \in \mathbb{R}}. This indicates x is a real number within the range -4 to just less than 3.

(ii) The intersection A ∩ B represents the numbers that are common to both sets A and B. Therefore, A ∩ B = {x : -2 \leq x < 3}. This tells us that x is a real number between -2 and just under 3.

For A ∩ B’, we are looking for numbers that are in A but not in B. Hence, A ∩ B’ = {x : 3 \leq x < 5}. Here, x is a real number that starts at 3 and goes up to, but does not include, 5.

Question 13

Given A = {x : -1 < x ≤ 5, x ∈ R} and B = {x : -4 ≤ x < 3, x ∈ R}

Represent on different number lines :

(i) A ∩ B

(ii) A’ ∩ B

(iii) A – B

Given sets are A = {x : -1 < x \leq 5, x \in R} and B = {x : -4 \leq x < 3, x \in R}.

(i) A ∩ B: This represents the numbers that are present in both sets A and B. For this intersection, we have:

The solution on the number line is depicted as:

(ii) A’ ∩ B: Here, we look for numbers that are not in set A but are in set B. Thus, we have:

The number line representation is:

(iii) A – B: This set includes numbers that are in A but not in B. Therefore, we find:

The corresponding number line is shown as:

Question 14

Find the range of values of x, which satisfy :

Graph, in each of the following cases, the values of x on different real number lines:

(i) x ∈ W

(ii) x ∈ Z

(iii) x ∈ R

We start with the inequality:

First, consider the left side of the inequality:

This can be rewritten as:

Subtract \dfrac{5}{3} from both sides:

Multiplying both sides by 2 gives:

Now, solve the right side of the inequality:

This simplifies to:

Subtract \dfrac{5}{3} from both sides:

Multiplying both sides by 2 gives:

Combining results from (i) and (ii), we have:

(i) When x \in W:

The solution set is {0, 1, 2, 3, 4, 5, 6}.

The representation on the number line is:

(ii) When x \in Z:

The solution set is {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}.

The representation on the number line is:

(iii) When x \in R:

The solution set is {x : -4 \le x < 7, x \in R}.

The representation on the number line is:

Question 15

Given : A = {x : -8 < 5x + 2 ≤ 17, x ∈ I}

B = {x : -2 ≤ 7 + 3x < 17, x ∈ R}

Where R = {real numbers} and I = {integers}

Represent A and B on two different numbers lines. Write down elements of A ∩ B.

Let’s start by solving the set A: A = \{x : -8 < 5x + 2 \leq 17, x \in I\}.

First, tackle the left-hand side:

⇒ 5x > -8 - 2

⇒ 5x > -10

⇒ x > -2 ……(i)

Next, focus on the right-hand side:

⇒ 5x \leq 17 - 2

⇒ 5x \leq 15

⇒ x \leq 3 …….(ii)

From inequalities (i) and (ii), we derive:

Since x \in I, the integers satisfying this condition are:

∴ Solution set = {-1, 0, 1, 2, 3}

Now, consider the set B: B = \{x : -2 \leq 7 + 3x < 17, x \in R\}.

Begin with the left-hand side:

⇒ 3x \geq -2 - 7

⇒ 3x \geq -9

⇒ x \geq -3 …….(iii)

Then, handle the right-hand side:

⇒ 3x < 17 - 7

⇒ 3x < 10

⇒ x < \dfrac{10}{3} ……..(iv)

Combining (iii) and (iv), we find:

The intersection A \cap B consists of elements common to both sets:

Hence, ( A \cap B ={-1, 0, 1, 2, 3}.

Question 16

Solve the following inequation and represent the solution set on the number line 2x – 5 ≤ 5x + 4 < 11, where x ∈ I.

Consider the inequation:

2x – 5 ≤ 5x + 4 < 11

First, let’s handle the left part:

⇒ 2x – 5 ≤ 5x + 4

Subtract 2x from both sides:

⇒ 5x – 2x ≥ -5 – 4

⇒ 3x ≥ -9

Divide by 3:

⇒ x ≥ -3 \, \text{….(i)}

Now, focus on the right part:

⇒ 5x + 4 < 11

Subtract 4 from both sides:

⇒ 5x < 11 – 4

⇒ 5x < 7

Divide by 5:

⇒ x < \dfrac{7}{5} \, \text{….(ii)}

Combine results from (i) and (ii):

-3 ≤ x < \dfrac{7}{5}

Since x is an integer (x ∈ I), we select integer values:

∴ Solution set = {-3, -2, -1, 0, 1}.

The solution is depicted on the number line as follows:

Test Yourself

Question 1(a)

The maximum value of x for the inequation 4x ≤ 12 + x is :

• (a) 5
• (b) 4
• (c) 3
• (d) 2.4

We start with the inequality:

⇒ 4x ≤ 12 + x

To isolate terms involving x, subtract x from both sides:

⇒ 4x – x ≤ 12

This simplifies to:

⇒ 3x ≤ 12

Next, divide each side by 3 to solve for x:

⇒ x ≤ \dfrac{12}{3}

Which further simplifies to:

⇒ x ≤ 4

∴ The largest possible value for x is 4.

Hence, Option 2 is the correct option.

Question 1(b)

The minimum value of x for the inequation 5x – 4 ≥ 18 – 6x is :

• (a) 2
• (b) 22
• (c) -22
• (d) -2

Starting with the inequality:

⇒ 5x – 4 ≥ 18 – 6x

First, rearrange the terms to bring all x terms to one side:

⇒ 5x + 6x ≥ 18 + 4

This simplifies to:

⇒ 11x ≥ 22

Now, divide both sides by 11:

⇒ x ≥ \dfrac{22}{11}

⇒ x ≥ 2

∴ The smallest possible value for x is 2.

Hence, Option 1 is the correct option.

Question 1(c)

If 1 ≤ -x < 5, x is an integer then the sum of smallest and greatest values of x is :

• (a) -5
• (b) -4
• (c) -3
• (d) 0

We have the inequality:

⇒ 1 ≤ -x < 5

To solve for x, reverse the inequality signs by multiplying through by -1:

⇒ -5 < x ≤ -1

Since x must be an integer, the possible integer values are: {-4, -3, -2, -1}.

The smallest integer in this set is -4 and the largest is -1.

Thus, the sum of the smallest and greatest values is:

-4 + (-1) = -5

Hence, Option 1 is the correct option.

Question 1(d)

The value of x for the inequation 3x + 15 < 5x + 13, x ∈ Z is :

• (a) > 1
• (b) < 1
• (c) = 1
• (d) ≥ 1

Let’s consider the inequality:

Rearranging the terms, we can express it as:

Subtracting 3x from both sides gives:

This simplifies to:

Dividing both sides by 2, we find:

Hence, Option 1 is the correct option.

Question 1(e)

The real number lines for two inequations A and B are as given below, A ∩ B is :

For the set A, we have the condition: -3 < x \leq 1, indicating that x is a real number greater than -3 and up to and including 1.

For the set B, the condition is: -4 \leq x < 0, meaning x is a real number starting from -4 up to but not including 0.

To determine the intersection A \cap B, we need the values that satisfy both conditions simultaneously.

Notice that the overlap of these two conditions is when x is greater than -3 but less than 0. Thus, A \cap B = \{x : x \in \mathbb{R} \text{ and } -3 < x < 0\}.

Hence, Option 1 is the correct option.

Question 1(f)

For the inequations A and B [as given above in part (d)], A ∪ B is :

For set A, we have all real numbers x such that -3 < x \leq 1. This means x is greater than -3 and up to 1.

For set B, x includes all real numbers where -4 \leq x < 0. This indicates x starts from -4 and goes up to, but does not include, 0.

The union A \cup B combines all elements from both sets A and B. Thus, A \cup B = \{x : x \in \mathbb{R} \text{ and } -4 \leq x \leq 1\}, covering numbers from -4 to 1 inclusive.

Hence, Option 1 is the correct option.

Question 1(g)

-\dfrac{3}{2} \le -\dfrac{2x}{3} where x ∈ R.

Assertion (A): The largest value of x is \dfrac{9}{4}.

Reason (R): When the signs of both the sides of an inequalities are changed, the sign of inequality reverses.

• (a) A is true, R is false.
• (b) A is false, R is true.
• (c) Both A and R are true and R is the correct reason for A.
• (d) Both A and R are true and R is the incorrect reason for A.

Both the assertion and the reason are accurate, and the reason correctly explains the assertion.

Reasoning

Let’s examine the assertion:

Thus, the assertion is valid as the largest possible value for x is indeed \dfrac{9}{4}.

Now, considering the reason:

When we multiply or divide both sides of an inequality by a negative number, the inequality sign reverses to ensure the inequality remains true.

Therefore, the reason is also correct.

Question 1(h)

Inequation 5 – 2x ≥ x – 10, where x ∈ N (Natural numbers)

Assertion (A): 5 – 2x ≥ x – 10 ⇒ -3x ≥ -15 ⇒ x ≥ 5

∴ Solution set = {5, 6, 7, 8, ……….}

Reason (R): 5 – 2x ≥ x – 10 ⇒ 5 + 10 ≥ 3x ⇒ x ≤ 5

∴ Solution set = {1, 2, 3, 4, 5}

• (a) A is true, R is false.
• (b) A is false, R is true.
• (c) Both A and R are true and R is correct reason for A.
• (d) Both A and R are true and R is incorrect reason for A.

Assertion (A) is incorrect, whereas Reason (R) holds true.

Explanation

Let’s analyze the Assertion:

The given inequation is 5 – 2x ≥ x – 10.

⇒ 5 – 2x + 10 ≥ x

⇒ -2x + 15 ≥ x

⇒ 15 ≥ x + 2x

⇒ 15 ≥ 3x

⇒ x ≤ \dfrac{15}{3}

⇒ x ≤ 5

∴ The solution set derived from this is {1, 2, 3, 4, 5}.

Thus, the Assertion (A) stating x ≥ 5 is incorrect.

Now, considering the Reason:

The solution set provided is {1, 2, 3, 4, 5}, which matches our derived solution set.

Thus, the Reason (R) is true.

Hence, A is false, R is true.

Question 1(i)

x ∈ W, x ≥ -3 and x < 5.

Statement (1) : There will be no solution for the given inequalities.

Statement (2) : The real number line for the given inequations is :

• (a) Both the statements are true.
• (b) Both the statements are false.
• (c) Statement 1 is true, and statement 2 is false.
• (d) Statement 1 is false, and statement 2 is true.

Statement 1 is false, and statement 2 is true.

Consider the inequality x \geq -3.

The solution set for this inequality is \{-3, -2, -1, 0, 1, 2, \ldots\}, indicating all whole numbers starting from -3 and increasing. Let’s label this as equation (1).

Now, examine the inequality x < 5.

The solution set for this is \{\ldots, 1, 2, 3, 4\}, which includes all whole numbers less than 5. We’ll call this equation (2).

By combining the results from (1) and (2), we find that the common solution set is \{-3, -2, -1, 0, 1, 2, 3, 4\}.

Thus, statement 1 is incorrect because there are indeed solutions.

Looking at the number line representation for these inequalities:

!x ∈ W, x \geq -3 and x < 5. Linear Inequations, Concise Mathematics Solutions ICSE Class 10.(https://www.icseboard.org/images/q1h-inequality-concise-maths-solutions-icse-class-10-1200×92.png)

This confirms that the number line accurately represents the solution set, so statement 2 is true.

Therefore, option 4 is correct.

Question 1(j)

5 + x ≤ 2x < x – 2, x ∈ R.

Statement (1) : There is no value of x ∈ R that satisfies the given inequation.

Statement (2) : 5 + x – x ≤ 2x – x < x – 2 – x ⇒ 5 ≤ x < -2

• (a) Both the statements are true.
• (b) Both the statements are false.
• (c) Statement 1 is true, and statement 2 is false.
• (d) Statement 1 is false, and statement 2 is true.

Both statements hold true.

Consider the inequation:

Given: 5 + x ≤ 2x < x – 2

Breaking it down, we have:

⇒ 5 + x ≤ 2x

⇒ 5 ≤ 2x – x

⇒ 5 ≤ x ………. (1)

Simultaneously, consider:

2x < x – 2

⇒ 2x – x < -2

⇒ x < -2 ………. (2)

Notice that no real number can simultaneously satisfy being both greater than or equal to 5 and less than -2.

∴ Both statements are indeed true.

Hence, option 1 is correct.

Question 2

Solve the inequation :

12 + 1\dfrac{5}{6}x \le 5 + 3x and x ∈ R.

Let’s start by considering the inequality:

Convert the mixed fraction to an improper fraction:

Rearrange terms to bring all terms involving x to one side and constants to the other:

Combine the x terms by finding a common denominator:

Simplify the fraction:

Clear the fraction by multiplying both sides by 6:

Simplifying gives:

∴ Solution set = {x : x ∈ R and x ≥ 6}.

Question 3

Given x ∈ {whole numbers}, find the solution set of :

-1 ≤ 3 + 4x < 23

Let’s begin by examining the given inequality:

-1 ≤ 3 + 4x < 23

First, address the left side of the inequality:

⇒ -1 ≤ 3 + 4x

Subtract 3 from both sides:

⇒ 4x ≥ -1 – 3

⇒ 4x ≥ -4

Divide by 4:

⇒ x ≥ -1 \quad \text{…(i)}

Now, consider the right side of the inequality:

⇒ 3 + 4x < 23

Subtract 3 from both sides:

⇒ 4x < 23 – 3

⇒ 4x < 20

Divide by 4:

⇒ x < 5 \quad \text{…(ii)}

Combining the results from (i) and (ii), we find:

-1 ≤ x < 5

Since x belongs to the set of whole numbers, the possible values for x are:

∴ Solution set = {0, 1, 2, 3, 4}.

Question 4

Find the set of values of x, satisfying :

7x + 3 ≥ 3x – 5 and \dfrac{x}{4} - 5 \le \dfrac{5}{4} - x, where x ∈ N.

First, let’s tackle the inequality:

Rearranging the terms, we have:

This simplifies to:

Dividing both sides by 4 gives:

This is our first inequality: (i)

Next, consider the second inequality:

Adding x to both sides, we get:

Combine the terms on the left:

Which simplifies to:

Multiplying both sides by \frac{4}{5}, we find:

This is our second inequality: (ii)

Combining (i) and (ii), we have:

Since x \in \mathbb{N}, the natural numbers that satisfy this are:

∴ Solution set = {1, 2, 3, 4, 5}.

Question 5

Solve :

(i) \dfrac{x}{2} + 5 \le \dfrac{x}{3} + 6, where x is a positive odd integer

(ii) \dfrac{2x + 3}{3} \ge \dfrac{3x - 1}{4}, where x is a positive even integer

(i) Begin by isolating terms:

Subtract \dfrac{x}{3} and 5 from both sides:

Combine the fractions:

Simplify:

Multiply through by 6:

Given that x is a positive odd integer, we have:

∴ Solution set = {1, 3, 5}.

(ii) Let’s handle the second inequality:

Subtract \dfrac{3x - 1}{4} from both sides:

Find a common denominator and simplify:

Simplify the numerator:

Combine like terms:

Rearrange to solve for x:

Since x is a positive even integer, we conclude:

∴ Solution set = {2, 4, 6, 8, 10, 12, 14}.

Question 6

Solve the inequation :

-2\dfrac{1}{2} + 2x \le \dfrac{4x}{5} \le \dfrac{4}{3} + 2x, x ∈ W.

Graph the solution set on the number line.

Consider the inequation:

Let’s tackle the left part of the inequation:

\Rightarrow -2\dfrac{1}{2} + 2x \le \dfrac{4x}{5}
\Rightarrow -\dfrac{5}{2} + 2x \le \dfrac{4x}{5}
\Rightarrow 2x - \dfrac{4x}{5} \le \dfrac{5}{2}
\Rightarrow \dfrac{10x - 4x}{5} \le \dfrac{5}{2}
\Rightarrow \dfrac{6x}{5} \le \dfrac{5}{2}
\Rightarrow x \le \dfrac{5}{2} \times \dfrac{5}{6}
\Rightarrow x \le \dfrac{25}{12}
\Rightarrow x \le 2\dfrac{1}{12} ........(i)

Now, let’s solve the right part of the inequation:

\Rightarrow \dfrac{4x}{5} \le \dfrac{4}{3} + 2x
\Rightarrow \dfrac{4x}{5} - 2x \le \dfrac{4}{3}
\Rightarrow \dfrac{4x - 10x}{5} \le \dfrac{4}{3}
\Rightarrow -\dfrac{6x}{5} \le \dfrac{4}{3}
\Rightarrow \dfrac{6x}{5} \ge -\dfrac{4}{3}
\Rightarrow x \ge -\dfrac{4}{3} \times \dfrac{5}{6}
\Rightarrow x \ge -\dfrac{10}{9}
\Rightarrow x \ge -1\dfrac{1}{9} .......(ii)

Combining the results from (i) and (ii), we have:

Since x \in W, which represents whole numbers:

∴ Solution set = {0, 1, 2}.

The solution set on the number line is depicted as follows:

Question 7

Find three consecutive largest positive integers such that the sum of one-third of first, one-fourth of second and one-fifth of third is at most 20.

Consider three consecutive positive integers: x, x + 1, and x + 2.

The problem states that the total of one-third of the first, one-fourth of the second, and one-fifth of the third should not exceed 20.

This simplifies to:

To eliminate the fractions, find a common denominator:

Simplify the expression:

Combine like terms:

Subtract 39 from both sides:

Divide both sides by 47:

Calculate the division:

Since x is an integer, the largest possible value is x = 24. Therefore, the integers are x = 24, x + 1 = 25, and x + 2 = 26.

Hence, three consecutive numbers are 24, 25, and 26.

Question 8

Solve the following inequation and represent the solution set on the number line :

4x – 19 < \dfrac{3x}{5} - 2 \le -\dfrac{2}{5} + x, x ∈ R

Consider the given inequation:

4x – 19 < \dfrac{3x}{5} - 2 \le -\dfrac{2}{5} + x

Let’s tackle the left side of the inequality first:

Now, let’s solve the right side of the inequality:

Combining results from (i) and (ii), we have:

-4 ≤ x < 5

∴ Solution set = {x : -4 ≤ x < 5, x ∈ R}.

The solution is represented on the number line as follows:

Question 9(i)

Find the greatest value of x ∈ Z, so that :

-1 ≤ 3 + 4x < 23

We have the inequality:

-1 ≤ 3 + 4x < 23

Let’s tackle the left side of the inequality first:

⇒ -1 ≤ 3 + 4x

Subtract 3 from both sides:

⇒ -1 – 3 ≤ 4x

⇒ -4 ≤ 4x

Now, divide each side by 4:

⇒ \frac{-4}{4} ≤ x

⇒ -1 ≤ x

Thus, we have x ≥ -1 ….(1)

Now, consider the right side of the inequality:

⇒ 3 + 4x < 23

Subtract 3 from both sides:

⇒ 4x < 23 – 3

⇒ 4x < 20

Divide each side by 4:

⇒ x < \frac{20}{4}

⇒ x < 5 ….(2)

Combining results from (1) and (2), we find:

-1 ≤ x < 5

Since x is an integer (x ∈ Z), the possible integer values are:

x = {-1, 0, 1, 2, 3, 4}

Therefore, the greatest value of x is 4.

Question 9(ii)

If 7 ≥ -2x + 1 > -7 ; find the sum of greatest and smallest values of x ∈ I.

We are given the compound inequality:

7 ≥ -2x + 1 > -7

Let’s handle the left side of the inequality first:

⇒ 7 ≥ -2x + 1

Subtract 1 from both sides:

⇒ 7 – 1 ≥ -2x

⇒ 6 ≥ -2x

To isolate x, divide by -2 (remember to flip the inequality sign):

⇒ 2x ≥ -6

⇒ x ≥ \dfrac{-6}{2}

⇒ x ≥ -3 \quad \text{…(1)}

Now, solve the right side of the inequality:

⇒ -2x + 1 > -7

Subtract 1 from both sides:

⇒ -2x > -7 – 1

⇒ -2x > -8

Divide by -2 (flip the inequality sign):

⇒ 2x < 8

⇒ x < \dfrac{8}{2}

⇒ x < 4 \quad \text{…(2)}

Combining results from (1) and (2), we have:

-3 ≤ x < 4

Since x must be an integer (x ∈ I), the possible values are:

x = {-3, -2, -1, 0, 1, 2, 3}

Here, the greatest value of x is 3 and the smallest value is -3.

Thus, the sum of the greatest and smallest values is:

-3 + 3 = 0

Hence, sum of greatest and smallest values of x = 0.

Case Study Based Question

Question 1

Case study :
A teacher asked to Rohan to draw a triangle with following condition: The longest side of the triangle is 7 cm less than twice the shortest side and third side is 7 cm shorter than longest side. The perimeter of the triangle is atleast 84 cm.

Based on the above information, form a linear inequation and answer the following questions :

(i) What is the minimum length of the shortest side ?

(ii) What is the minimum length of the longest side ?

(iii) Identify the type of triangle that Rohan has drawn along with the length possible sides he got.

(iv) What is the least area of the triangle drawn ?

Assume the shortest side of the triangle is x cm.

According to the conditions given:

• The longest side measures 2x - 7 cm.
• The third side is 7 cm shorter than the longest side, hence it is (2x - 7) - 7 = 2x - 14 cm.

(i) With the perimeter being at least 84 cm, we have:

Simplifying gives:

Adding 21 to both sides:

Dividing by 5:

Thus, the shortest side must be at least 21 cm.

(ii) For the longest side, substituting x = 21:

Thus, the longest side must be at least 35 cm.

(iii) Calculating the third side when x = 21:

The sides of the triangle are 21 cm, 28 cm, and 35 cm.

Checking if it’s a right triangle:

Since 21^2 + 28^2 = 35^2, it confirms a right-angled triangle with the hypotenuse being 35 cm.

Hence, the triangle is right-angled.

(iv) Using the formula for the area of a right-angled triangle:

Calculating gives:

Therefore, the least area of the triangle is 294 cm^2.